square on a cubic: Find k so square has 4 corners on y = x3 − kx

[apple2357]

I have been playing around with this problem but not making much progress:

For which values of k is it possible to draw a square using four points that lie on the curve with equation y = x³ − kx?

Any hints/thoughts to get started would be gratefully received!

What i have tried to so far:

- I have thought about what properties might be true for four points to form a square:
For example, if ( a, b) is the coordinates of the first point say A, then B could be (a+c, b+d) and C (a+c-d, b+d+c) & D(a-d, b+c)

So a=3, b=4 and c=2, d=5 ( for example). The following points form a square:
A= (3,4) B= (5,9) C= (0,11) and D=(-2, 6)
Now do these lie on y= y = x³ − kx ?

It doesn't appear so as they don't solve for a unique value of k.

If i keep it general, the algebra gets unbearable. Is there some kind of symmetry argument i am missing?

 

[Steven G]

I have been playing around with this problem but not making much progress:

For which values of k is it possible to draw a square using four points that lie on the curve with equation y = x³ − kx?

Any hints/thoughts to get started would be gratefully received!

What i have tried to so far:

- I have thought about what properties might be true for four points to form a square:
For example, if ( a, b) is the coordinates of the first point say A, then B could be (a+c, b+d) and C (a+c-d, b+d+c) & D(a-d, b+c)

So a=3, b=4 and c=2, d=5 ( for example). The following points form a square:
A= (3,4) B= (5,9) C= (0,11) and D=(-2, 6)
Now do these lie on y= y = x³ − kx ?

It doesn't appear so as they don't solve for a unique value of k.

If i keep it general, the algebra gets unbearable. Is there some kind of symmetry argument i am missing?

I'll think about this. I do think that 1st picking the points is not a good idea.

 

[Steven G]

I have been playing around with this problem but not making much progress:

For which values of k is it possible to draw a square using four points that lie on the curve with equation y = x³ − kx?

Any hints/thoughts to get started would be gratefully received!

What i have tried to so far:

- I have thought about what properties might be true for four points to form a square:
For example, if ( a, b) is the coordinates of the first point say A, then B could be (a+c, b+d) and C (a+c-d, b+d+c) & D(a-d, b+c)

So a=3, b=4 and c=2, d=5 ( for example). The following points form a square:
A= (3,4) B= (5,9) C= (0,11) and D=(-2, 6)
Now do these lie on y= y = x³ − kx ?

It doesn't appear so as they don't solve for a unique value of k.

If i keep it general, the algebra gets unbearable. Is there some kind of symmetry argument i am missing?

What point do you think that the center of the square will be? Can you prove it?

 

[JeffM]

apple, I have not worked out this approach, but give it a try. It appears promising to me.

First, start with four points as you did, but take advantage of the fact that you know the relationship between the x and y values. That is points A, B, C, and D have respective co-ordinates of

\(\displaystyle (a, \ a^3 - ak),\ (b, \ b^3 - bk),\ (c,\ c^3 - ck), \text { and } (d,\ d^3 - dk).\)

Now you can form linear functions for the lines forming the sides. The slopes of lines AB and CD will equal, and the slopes of BC and AD will equal each other and will equal the negative reciprocals of the slopes of the other two lines. Furthermore, the distances between the points are all equal.

So you get a bunch of equations (presumably many of them are not independent). If there are any constraints on k, they ought to become apparent when you try to determine a, b, c, d, and k.

 

[apple2357]

What point do you think that the center of the square will be? Can you prove it?

Not sure,but i will play with this. Will the centre of the square be at the origin ( point of inflexion?)

 

[apple2357]

apple, I have not worked out this approach, but give it a try. It appears promising to me.

First, start with four points as you did, but take advantage of the fact that you know the relationship between the x and y values. That is points A, B, C, and D have respective co-ordinates of

\(\displaystyle (a, \ a^3 - ak),\ (b, \ b^3 - bk),\ (c,\ c^3 - ck), \text { and } (d,\ d^3 - dk).\)

Now you can form linear functions for the lines forming the sides. The slopes of lines AB and CD will equal, and the slopes of BC and AD will equal each other and will equal the negative reciprocals of the slopes of the other two lines. Furthermore, the distances between the points are all equal.

So you get a bunch of equations (presumably many of them are not independent). If there are any constraints on k, they ought to become apparent when you try to determine a, b, c, d, and k.

I will try this too!

A bit later.. algebraically this feels heavy and whilst some things are cancelling, there is a lot of terms to keep an eye on!

 

[apple2357]

Been playing on dynamic software for inspiration... this doesn't quite work:

 

[apple2357]

More messing around and my conjecture is that the centre of the square lies on the curve. Not sure exactly why and this 'looks' like it works!
Of course, it is not a proof.
I am starting to wonder whether it is ALWAYS possible..

 

[Steven G]

More messing around and my conjecture is that the centre of the square lies on the curve. Not sure exactly why and this 'looks' like it works!
Of course, it is not a proof.
I am starting to wonder whether it is ALWAYS possible..

If the center of the is located at the origin, then what would the 3 corner points of the square look like? The answer to this question might make this problem a bit easier!

 

[Dr.Peterson]

I am reasonably certain, due to symmetry, that the center of the square must lie at the origin.

On GeoGebra, I put the center there and constructed a square with one vertex at a point A on the curve, with the other three vertices obtained by rotation about the center. This led to some beautiful discoveries. In particular, consider what path the adjacent vertex, B, follows as you move A along the curve. You'll see what is required of the curve in order to have at least one solution.

I'm not sure yet whether that condition can be derived algebraically, as that attempt led to a 9th degree equation (which reduced to a 4th degree, but no more). Note that we don't necessarily have to be able to find the point A algebraically, though; we just need to find k such that it is possible (which may involve a discriminant).

 

[apple2357]

If the center of the is located at the origin, then what would the 3 corner points of the square look like? The answer to this question might make this problem a bit easier!

Sorry, still too cryptic for me!

 

[Steven G]

Sorry, still too cryptic for me!

I meant 4 corners of the square. YOU listed the 4 points in an earlier post, now that you know that the center of the square is at the origin can you update those 4 points? Hint: The diagonals of a rectangle passes through the center of the rectangle.

 

[apple2357]

I meant 4 corners of the square. YOU listed the 4 points in an earlier post, now that you know that the center of the square is at the origin can you update those 4 points? Hint: The diagonals of a rectangle passes through the center of the rectangle.

Ok, i have established that the following four points lie on a square ( where the centre is at the origin). This certainly reduces the algebra for the next stage i think...

A= (a, b)
B= (-b, a)
C= (-a,-b)
D= (b, -a)

I have verified it on GGB using sliders:

 

[apple2357]

Ok... a little while later

It appears that if we use the coordinates above in y = x^3 - k x

We get

ka = a^3 -b
kb = a+ b^3

With some manipulation it follows that if (a,b) satisfies the cubic equation ab^3+b^2-a^3b-a^2=0 then
k = (a(a^2+1) + b(b^2-1))/(a+b). I can't seem to find a way to simplify this.
And then the curve passes through four points of a square!


For example, if a= 1 , b= -1.84 and k =2.84

Or if a=2 , b= -2.46, 0.6, 1.36 then k = 5.28, 3.7 and 3.32 respectively.

Screenshots below if you don't believe me!

Two remaining questions for me

1. How can i be sure the centre of the square passes through the origin?
2. Can the above algebra be simplified?

 

[Dr.Peterson]

It appears that if we use the coordinates above in y = x^3 - k x

We get

ka = a^3 -b
kb = a+ b^3

With some manipulation it follows that if (a,b) satisfies the cubic equation ab^3+b^2-a^3b-a^2=0 then
k = (a(a^2+1) + b(b^2-1))/(a+b). I can't seem to find a way to simplify this.
And then the curve passes through four points of a square!

For example, if a= 1 , b= -1.84 and k =2.84

Or if a=2 , b= -2.46, 0.6, 1.36 then k = 5.28, 3.7 and 3.32 respectively.

Two remaining questions for me

1. How can i be sure the centre of the square passes through the origin?
2. Can the above algebra be simplified?

I don't have a proof that the center must be at the origin; it just seems "obvious", or at least it is a good way to make the work easier through symmetry.

What you've done is to find that if there is such a square, then the vertices (a,b) all lie on the quartic equation (in two variables) ab^3+b^2-a^3b+a^2=0, where I have corrected a sign. (See the graph below.) Thus, they are among the 8 intersections of that curve with the original. This doesn't help us much toward the goal of finding the possible values of k. (Your equation for k requires already knowing a and b, so it doesn't take us anywhere.)

What I did is nearly equivalent (and similarly not yet helpful). Rather than rotate the given vertex to find the other three and use them as you did, I rotated the cubic curve itself (because two vertices will lie on the original, and the other two on the rotated curve). In the graph below, you can see that the vertices therefore lie on the (possibly 8) intersections of the two cubics (blue and green), which are also where your quartic (cyan broken line) intersects the cubic. As I vary k, I find that for k =1.84 or so, the cubics just touch, and for anything less, they don't (except at the origin), so there is no square possible. This (if we could determine that number directly) is the answer to the problem. The picture is for k=3.



Note that when you combine the equations y = x^3 - kx and -x = y^3 -ky (the rotated one), you get a 9th degree equation for the intersections, which agrees with the picture (9 intersections at most). In effect, we would like to use some sort of discriminant to determine when the 8 intersections exist, in order to solve the problem. I can get the equation down to a 4th degree (in x^2), but that's where I'm stopped.

 

[apple2357]

I don't have a proof that the center must be at the origin; it just seems "obvious", or at least it is a good way to make the work easier through symmetry.

What you've done is to find that if there is such a square, then the vertices (a,b) all lie on the quartic equation (in two variables) ab^3+b^2-a^3b+a^2=0, where I have corrected a sign. (See the graph below.) Thus, they are among the 8 intersections of that curve with the original. This doesn't help us much toward the goal of finding the possible values of k. (Your equation for k requires already knowing a and b, so it doesn't take us anywhere.)

What I did is nearly equivalent (and similarly not yet helpful). Rather than rotate the given vertex to find the other three and use them as you did, I rotated the cubic curve itself (because two vertices will lie on the original, and the other two on the rotated curve). In the graph below, you can see that the vertices therefore lie on the (possibly 8) intersections of the two cubics (blue and green), which are also where your quartic (cyan broken line) intersects the cubic. As I vary k, I find that for k =1.84 or so, the cubics just touch, and for anything less, they don't (except at the origin), so there is no square possible. This (if we could determine that number directly) is the answer to the problem. The picture is for k=3.

View attachment 9802

Note that when you combine the equations y = x^3 - kx and -x = y^3 -ky (the rotated one), you get a 9th degree equation for the intersections, which agrees with the picture (9 intersections at most). In effect, we would like to use some sort of discriminant to determine when the 8 intersections exist, in order to solve the problem. I can get the equation down to a 4th degree (in x^2), but that's where I'm stopped.

Can the 8th degree equation be expressed in the form (x²-p²)²(x²-q²)² to give us something? Which is what i think you mean in your last sentence?

 

[Dr.Peterson]

Can the 8th degree equation be expressed in the form (x²-p²)²(x²-q²)² to give us something? Which is what i think you mean in your last sentence?

No, that would only have four solutions (p, -p, q, -q).

The equation I ended up with has the form ax^8 + bx^6 + cx^4 + dx^2 + e. It would factor something like (x^2 - p)(x^2 - q)(x^2 - r)(x^2 - s) so that there are four pairs of solutions. But those will be functions of k, and for k less than my 1.84 they will not exist (perhaps p,q,r,s would be negative or even imaginary).

 

[apple2357]

No, that would only have four solutions (p, -p, q, -q).

The equation I ended up with has the form ax^8 + bx^6 + cx^4 + dx^2 + e. It would factor something like (x^2 - p)(x^2 - q)(x^2 - r)(x^2 - s) so that there are four pairs of solutions. But those will be functions of k, and for k less than my 1.84 they will not exist (perhaps p,q,r,s would be negative or even imaginary).

Ok I have something like this:

x^8 - 3kx^6 -3k^2x^4+ (k^3+k)x^2 - k^2-1 =0

Is that what we are talking about?
For what values of k will this have 4 solutions? But won't the solutions be symmetric because it is an even function?

 

[Dr.Peterson]

Ok I have something like this:

x^8 - 3kx^6 -3k^2x^4+ (k^3+k)x^2 - k^2-1 =0

Is that what we are talking about?
For what values of k will this have 4 solutions? But won't the solutions be symmetric because it is an even function?

That looks like what I got; I have a couple signs different, but I had reason to think there might be a mistake, so you're likely right. (I never trust myself 100% until I have an answer I can check!)

Yes, the solutions (for x) are symmetric. Just look at my graph! One set of four will correspond to one square, and the rest to a second square. When k is right on the edge (i.e. about 2.84), there will be only four solutions; maybe that's the thing to look for: maybe we can find a value of k for which the equation factors as the product of four squares (that is, the square of a quartic). And maybe plugging in k=2.84 (as a check) will lead to some useful discovery.

Go to it, and I'll follow along as I have time.

 

[apple2357]

Still struggling on this...I tried letting u= x^2, reducing to a quartic and then thinking about (u^2+au+b)(u^2+cu+d) and trying to compare coefficients but that doesn't go anywhere nice!